Two characterizations of finite nilpotent groups

Theorem A.

Let G be a finite group. Then G is not p-nilpotent if and only if there are two elements g,h∈G with o⁢(g)=o⁢(h)=qk for some prime q≠p and

Theorem 1.1.

A finite group G is solvable if and only if μG,w⁢(1) is bounded away from zero as w ranges over all words.

Theorem B.

A word w∈F⁢[x,y] is in P if and only if the length of w is greater than 1 and for any abelian group G with order less than or equal to the length of w, the structure (G,*w) is a quasigroup.

Lemma 1.2.

If G is a nilpotent group and w∈P, then μG,w⁢(g)=1|G| for all g∈G, i.e., μG,w is uniform.

Theorem C.

A finite group G is nilpotent if and only if for all w in P with length less than 4⁢|G|, we have μG,w⁢(1)=1|G|.

Theorem 2.1.

Let G be a finite group, and suppose p is a prime. Then the following are equivalent:

1. G has a normal p-complement.

2. 𝐍G⁢(X) has a normal p-complement for every nonidentity p-subgroup X⊆G.

3. 𝐍G⁢(X)/𝐂G⁢(X) is a p-group for every p-subgroup X⊆G.

Lemma 2.2 ([3, Exercise 4D.4]).

Let A act via automorphisms on G, where G is a p-group and (|G|,|A|)=1. Suppose that A acts trivially on every A-invariant proper subgroup of G, but that the action of A on G is nontrivial. Then the exponent of G is p or 4.

Theorem 2.3.

Let G be a finite group. Then G is not p-nilpotent if and only if there are two elements g,h in G with o⁢(g)=o⁢(h)=qk for some prime q≠p and

Proof.

Clearly the existence of two such elements implies that G is not p-nilpotent. Suppose now that G is not p-nilpotent. Since G does not have a normal p-complement for the prime p, by the Frobenius Complement Theorem, there is a p-subgroup H<G, such that |𝐍G(H):𝐂G(H)| is divisible by a prime q≠p. We can assume H is minimal with this property, i.e., for all p-subgroups K<H, the groups 𝐍G⁢(K)/𝐂G⁢(K) are p-groups. Let Q be a Sylow q-subgroup of 𝐍G⁢(H). Then Q acts on H via automorphisms nontrivially, and by the minimality of H, we see that Q centralizes every Q-invariant subgroup of H. By Lemma 2.2 the exponent of H is either p or 4.

Let x∈H such that for some t∈Q we have xt≠x. Then

Since Q normalizes H, xt⁢x-1∈H and has order either p or 4. The element t is in Q and thus o⁢(t-1)=o⁢(x⁢t⁢x-1)=qk. ∎

Lemma 3.1.

Let G be a group and let Z=Z⁢(G). Let w∈P. The structure (G,*w) is a quasigroup if and only if (G/Z,*w) is a quasigroup.

Proof.

Suppose that (G,*w) is a quasigroup. Then w⁢(a,b)=w⁢(a,c) implies b=c. Use the “overbar notation” for the correspondence between elements of G and G/Z and suppose w⁢(a¯,b¯)=w⁢(a¯,c¯). Then

and we conclude that w⁢(a,b)=w⁢(a,c)⁢z for some z∈Z. Since w∈𝒫, we see that w⁢(a,c)⁢z=w⁢(a,c⁢z±1) and thus b¯=c¯. By symmetry, w⁢(a¯,b¯)=w⁢(c¯,b¯) implies that a¯=c¯ and (G/Z,*w) is a quasigroup.

Suppose that (G/Z,*w) is a quasigroup and w⁢(a,b)=w⁢(a,c). Then b¯=c¯ and we conclude that c=b⁢z for some z∈Z. But

and we conclude that z=1 and thus c=b. By symmetry, we see that (G,*w) is a quasigroup. ∎

Corollary 3.2.

Let G be a nilpotent group and let w∈P. Then (G,*w) is a quasigroup.

Theorem 3.3.

A word w∈F⁢[x,y] is in P if and only if the length of w is greater than 1 and for any abelian group G with order less than or equal to the length of w, the structure (G,*w) is a quasigroup.

Proof.

As stated in the above corollary to Lemma 3.1, for any nilpotent group G and any word w in 𝒫 the structure (G,*w) is a quasigroup.

Now suppose that w is not in 𝒫 and the length of w is greater than 1. We will demonstrate that there is a cyclic group G, with order less than or equal to the length of w, for which (G,*w) is not a quasigroup. Without loss of generality, either Totw⁢(x)=0 or there is a prime p that divides Totw⁢(x). If Totw⁢(x)=0, then for any abelian group g*1=1 for all g∈G, and we have that (C2,*w) is not a quasigroup. Suppose Totw⁢(x) is divisible by p. Then consider Cp, the cyclic group of order p, and note that g*1=1 for all g∈G. Therefore (Cp,*w) is not a quasigroup. Note that either Totw⁢(y)=0 (and hence (C2,*w) is not a quasigroup), or p is strictly less than the length of w. ∎

Theorem 3.4.

A finite group G is nilpotent if and only if for all w in P with length less than 4⁢|G|, the function μG,w⁢(g)=1|G| for all g∈G.

Proof.

For a nilpotent group G the structure (G,*w) is a quasigroup; therefore μG,w⁢(g)=1|G| for all g∈G. Suppose that G is not nilpotent. We will construct a word w, so that (G,*w) is not a quasigroup. In particular, suppose that G is not p-nilpotent for the prime p. Then by Theorem A there are two elements g and h in G of order qk, for a prime q distinct from p, such that o⁢(g⁢h)=p or 4. We will abuse notation and let p=o⁢(g⁢h). Let 0<a<qk be an inverse of p modulo qk, that is a⁢p≡1(modqk). Since (a,qk)=1, there are unique g^ and h^ such that g^a=g and h^a=h. Moreover, o⁢(g^)=o⁢(h^)=qk. Let b satisfy a⁢p-b⁢qk=1.

If o⁢(g⁢h)=p, consider the word

We note the following observations about w⁢(x,y):

1. Totw⁢(x)=Totw⁢(y)=a⁢p-b⁢qk=1. Hence w⁢(g,g-1)=1 for all g∈G.

2. w⁢(g^,h^)=(g⁢h)p-1⁢(g^a-b⁢qk⁢h^a-b⁢qk)=(g⁢h)p=1.

Since g^ is not h^-1 (since g≠h-1), we conclude that μG,w⁢(1)>1|G|.

The length of w is 2⁢(a⁢p+b⁢qk). We note that a<qk and b⁢qk=a⁢p-1<|G|. Hence the length of w is less than 4⁢|G|. ∎